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Post by principled on Feb 14, 2011 13:37:20 GMT 1
Setting the scene. Last night I was sitting looking through a doorway at a window in an adjoining room. It was dark outside and I could see the reflection of the television in the window. I went outside an looked into the room where the TV was and obviously was able to watch the programme as well.
Question: Why are some of the photons from the TV picture reflected by the window whilst some pass through it to the outside? Is the intensity of the image less when looked at through the glass due to the photons that have been reflected (I assume so)? Does the amount of light coming in (eg daylight vs night) have any effect on the amount of light from the TV that is reflected by the glass? P
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Post by Progenitor A on Feb 14, 2011 13:52:43 GMT 1
Feynman lists this very phenomenon as on eof th ethings that we accept as normal but have not a clue as to why some photons go through an some are reflected
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Post by speakertoanimals on Feb 14, 2011 17:34:10 GMT 1
I think the question wasn't actually about quantum stuff and why some photons are reflected and some transmitted, but WHY reflection and transmission in general. The first point is that the amount reflected doesn't depend on what other light may be passing through the glass, but just on the properties of the glass. You might not be able to see the reflection if theiir is strong light coming in from outside, but that just because your eyes can't distinguish it, not because it isn't there! We first need to look at what happens when an em wave encounters a material (like glass). Let's take a metal for starters. The em wave, being oscillating electric and megnetic fields, tugs on the charges in the material -- these may be the conduction electrons floating about in metals, or they may be the charges within molecules etc. The em wave causes the charges to move, and then, being moving charges, these re-radiate, in the classical approximation. The net effect is hence the incident wave plus the effect of the waves radiated by all the moving charges in the medium. When we are looking at refraction, and the speed of waves changing in materials, we look at the refractive index, which takes all those complicated interactions above, and packages up the net result as a change in speed of the wave, and a change in angle. But the same physical property also gives an idea of how much light is reflected. So, you take the difference in refractive indices, divide by the sum, and square it to get an estimate of the amount reflected when falling perpendicularly on the surface. THe Fresnel equations that describe the amount of transmission and reflection depend on solving for two classical em waves incident on a boundary between two media with different electrical and magnetic properties. The result depends on the exact polarisation of the incident wave. hyperphysics.phy-astr.gsu.edu/hbase/phyopt/freseq.htmlSo the case where we can treat light as classical waves (which takes the whole complicated structure of a material and replaces it with refractive index) is relatively straightforward. Note that it doesn't allow for cases where the material actually absorbs a significant amount of the incident radiation, so that reflection and transmission are the only options. It expains things such as total internal reflection, and why you can get polarised light by reflection. In quantum terms, we have two things to consider -- how the exact structure of the material translates into particular properties when subjected to incident radiation at a given frequency, and what happens at the photon level when a single photon hits the glass. In this sense, its like almost every other quantum physics experiment -- where there is more than one possible outcome, a given system does one or the other, with some associated probability. But at the root it is indeterminate -- two identical photons, one may be reflected, the other transmitted, and there is nothing that distinguishes them beforehand. God plays dice with the universe!
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